English
Maharashtra State BoardSSC (English Medium) 10th Standard

Prave That: √ 1 − Sin θ 1 + Sin θ = Sec θ − Tan θ

Advertisements
Advertisements

Question

Prove that:

\[\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \sec\theta - \tan\theta\]
Sum
Advertisements

Solution

\[\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}}\]
\[ = \sqrt{\frac{1 - \sin\theta}{1 + \sin\theta} \times \frac{1 - \sin\theta}{1 - \sin\theta}}\]
\[ = \sqrt{\frac{\left( 1 - \sin\theta \right)^2}{1 - \sin^2 \theta}}\]
\[ = \sqrt{\frac{\left( 1 - \sin\theta \right)^2}{\cos^2 \theta}} \left( \cos^2 \theta + \sin^2 \theta = 1 \right)\]

\[= \frac{1 - \sin\theta}{\cos\theta}\]

\[ = \frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}\]

\[ = \sec\theta - \tan\theta\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 6: Trigonometry - Practice Set 6.1 [Page 131]

APPEARS IN

Balbharati Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
Chapter 6 Trigonometry
Practice Set 6.1 | Q 6.03 | Page 131

RELATED QUESTIONS

If \[\cot\theta = \frac{40}{9}\], find the values of cosecθ and sinθ.


If tanθ = 1 then, find the value of

`(sinθ + cosθ)/(secθ + cosecθ)`


Prove that:

\[\sec\theta + \tan\theta = \frac{\cos\theta}{1 - \sin\theta}\]

Prove that:
If \[\tan\theta + \frac{1}{\tan\theta} = 2\], then show that \[\tan^2 \theta + \frac{1}{\tan^2 \theta} = 2\]


Choose the correct alternative answer for the following question.
cosec 45° =?


Choose the correct alternative answer for the following question.

When we see at a higher level, from the horizontal line, angle formed is ........
 

Prove the following.
(secθ + tanθ) (1 – sinθ) = cosθ


Prove the following.
sec2θ + cosec2θ = sec2θ × cosec2θ 


Prove the following.
cot2θ – tan2θ = cosec2θ – sec2θ


Prove the following:

sec6x – tan6x = 1 + 3sec2x × tan2x


Prove the following.

\[\frac{\tan\theta}{\sec\theta + 1} = \frac{\sec\theta - 1}{\tan\theta}\]


Choose the correct alternative: 
sinθ × cosecθ =?


Show that: 

`sqrt((1-cos"A")/(1+cos"A"))=cos"ecA - cotA"`


In ΔPQR, ∠P = 30°, ∠Q = 60°, ∠R = 90° and PQ = 12 cm, then find PR and QR.


Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`

= `((1 - square)/square) ((square + square)/(square  square))`

= `square/square xx 1/(square  square)`  ......`[(∵ square + square = 1),(∴ square = 1 - square)]`

 = `square/(square  square)`

= tan θ.sec θ

= R.H.S.

∴ L.H.S. = R.H.S.

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×